When should I use Integration by Parts versus u-substitution?

Use u-substitution when the integrand is a composite function and the derivative of the inner function appears in the integrand. Use Integration by Parts when you have a product of two functions where u-substitution does not directly apply, such as x e^x, x \sin(x), or x \ln(x).

How do I remember the Integration by Parts formula?

A common mnemonic is "U-dV equals U-V minus V-dU," which rhymes with the formula \int u\,dv = uv - \int v\,du. Another memory aid: think of it as reversing the product rule for derivatives, which says (uv)' = u'v + uv'.

What do I do if I choose u and dv incorrectly?

If your resulting integral \int v\,du is harder than the original, you made a poor choice. Try switching your assignments or select a different u entirely using the LIATE rule to guide you toward a better pairing.

Can I use Integration by Parts on \int e^{x^2}\,dx?

No. There is no elementary antiderivative for e^{x^2}, so Integration by Parts cannot help you evaluate it in closed form. This integral requires numerical methods or special functions like the error function.

Integration by Parts

Integration by Parts is a calculus technique for evaluating integrals of products by transforming them into simpler forms when substitution fails.

\int u\,dv = uv - \int v\,du

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What each symbol means

What Integration by Parts takes

Symbol	Meaning
$u$	One factor of the product in the integrand, chosen so that its derivative $du$ is simpler; if this choice makes the resulting integral $\int v\,du$ harder than the original, you chose incorrectly.
$v$	The antiderivative of the other factor; computed by integrating $dv$ and then multiplied by $u$ in the formula, so it must be found accurately.

When to use it

Use Integration by Parts when you have an integral of a product and u-substitution does not apply.

Level

Usually taught in: Calculus II · Appears on: AP Calculus

Worked examples

1. Integrate polynomial times sine

Problem

Find

\int x \sin(x)\,dx

$u = x, \quad dv = \sin(x)\,dx$
Choose the polynomial as $u$ because its derivative will be simpler than the original factor.
$du = dx, \quad v = -\cos(x)$
Differentiate $u$ to get $dx$ , and integrate $dv$ using $\int \sin(x)\,dx = -\cos(x)$ .
$\int x \sin(x)\,dx = x \cdot (-\cos(x)) - \int (-\cos(x))\,dx$
Substitute into the Integration by Parts formula $\int u\,dv = uv - \int v\,du$ .
$= -x\cos(x) + \int \cos(x)\,dx$
Simplify by distributing the negative sign.
$= -x\cos(x) + \sin(x) + C$
Integrate $\cos(x)$ to get $\sin(x)$ and add the constant of integration.

Answer: $-x\cos(x) + \sin(x) + C$

This problem pairs a polynomial with a trigonometric function, making it a prime candidate for Integration by Parts. We choose $u = x$ to reduce the degree under differentiation, leaving us with a simple integral of cosine to evaluate at the end.

2. Integrate polynomial times exponential with a negative exponent

Problem

Find

\int x e^{-2x}\,dx

$u = x, \quad dv = e^{-2x}\,dx$
Choose the polynomial part as $u$ so that differentiation simplifies it to a constant.
$du = dx, \quad v = -\frac{1}{2}e^{-2x}$
Differentiate $u$ to get $dx$ , and integrate $dv$ using the formula $\int e^{ax}\,dx = \frac{1}{a}e^{ax}$ with $a = -2$ .
$\int x e^{-2x}\,dx = x \cdot \left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right)\,dx$
Apply Integration by Parts: $\int u\,dv = uv - \int v\,du$ .
$= -\frac{1}{2}xe^{-2x} + \frac{1}{2}\int e^{-2x}\,dx$
Simplify and factor out the constant $\frac{1}{2}$ from the remaining integral.
$= -\frac{1}{2}xe^{-2x} + \frac{1}{2} \cdot \left(-\frac{1}{2}e^{-2x}\right) + C$
Integrate $e^{-2x}$ to get $-\frac{1}{2}e^{-2x}$ .
$= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C$
Simplify the arithmetic by multiplying the fractions.

Answer: $-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C$

This example involves negative exponents and fractions, requiring careful attention when computing $v$ and managing coefficients. The polynomial times exponential pattern appears frequently in applications, making Integration by Parts the natural choice when u-substitution cannot eliminate the polynomial factor.

3. Garden fencing cost problem

Problem

A rectangular garden plot needs a curved fence. The labor cost at distance

x

meters from the entrance is

(x+1)e^{-x}

dollars per meter. Find

\int (x+1)e^{-x}\,dx

to determine the cumulative cost function.

$u = x+1, \quad dv = e^{-x}\,dx$
Choose the linear polynomial $x+1$ as $u$ because differentiation eliminates it to a constant.
$du = dx, \quad v = -e^{-x}$
Differentiate $u$ to get $dx$ , and integrate $dv$ using $\int e^{-x}\,dx = -e^{-x}$ .
$\int (x+1)e^{-x}\,dx = (x+1) \cdot (-e^{-x}) - \int (-e^{-x})\,dx$
Apply the Integration by Parts formula.
$= -(x+1)e^{-x} + \int e^{-x}\,dx$
Simplify by distributing the negative and rewriting the integral with positive sign.
$= -(x+1)e^{-x} - e^{-x} + C$
Integrate $e^{-x}$ to get $-e^{-x}$ .
$= -e^{-x}(x+2) + C$
Factor out $-e^{-x}$ to write the answer in simplified form.

Answer: $-(x+1)e^{-x} - e^{-x} + C \text{ or } -e^{-x}(x+2) + C$

This real-world scenario shows how Integration by Parts models cumulative costs in practical settings. The product of a linear term and an exponential decay factor is typical in economics and engineering; integrating by parts cleanly separates the polynomial's contribution from the exponential's behavior.

Common mistakes

Where Integration by Parts usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\int u\,dv = uv + \int v\,du$ instead of $uv - \int v\,du$ .	The negative sign comes directly from the product rule: $(uv)' = u'v + uv'$ rearranges to $u'v = (uv)' - uv'$ , which becomes $\int u\,dv = uv - \int v\,du$ when integrated.	Always write the formula as $\int u\,dv = uv - \int v\,du$ and double-check the minus sign when substituting.
Choosing $u$ to be the exponential or trigonometric function (e.g., $u = e^x$ when the integrand is $x e^x$ ), then finding that $\int v\,du$ is much harder than the original integral.	The goal of Integration by Parts is to make the new integral $\int v\,du$ simpler than the original. Prioritizing $u$ incorrectly defeats the purpose and creates harder work.	Use the LIATE rule: prioritize Logarithmic, Inverse trig, Algebraic (polynomial), Trigonometric, Exponential. If the resulting integral is harder, stop and swap your choice of $u$ and $dv$ .
Forgetting to write $+ C$ at the end of an indefinite integral, or writing it during intermediate steps before integration is complete.	The constant of integration represents all antiderivatives and belongs only at the final answer. Including it mid-derivation suggests the integration is finished when it is not.	Omit $+ C$ in all intermediate steps and add it only to the final simplified result.

Tips and when to use something else

Remember the LIATE rule to choose $u$ : prioritize Logarithmic, Inverse trig, Algebraic (polynomial), Trigonometric, and Exponential functions in that order. Choose $u$ from whichever category your integrand contains first.
If the integral $\int v\,du$ is harder than the original, you chose $u$ and $dv$ backward. Stop and switch them, or try u-substitution on a different part of the integrand.
For integrals like $\int x^n e^x\,dx$ , use Integration by Parts; for integrals like $\int e^{x^2}\,dx$ (no elementary antiderivative), Integration by Parts cannot help—use numerical integration or the error function instead.
When Integration by Parts produces a cycle (e.g., applying it twice to $\int e^x \sin(x)\,dx$ returns a multiple of the original integral), rearrange the equation algebraically to solve for the unknown integral rather than applying Integration by Parts a third time.

Related formulas

Indefinite IntegralThe indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.Partial FractionsPartial Fractions breaks a rational function into a sum of simpler fractions, each of which is easier to integrate or analyze.Integral of e^xThe integral of e^x equals e^x plus C, the essential antiderivative for solving exponential growth, decay, and differential equations.Trigonometric SubstitutionTrigonometric substitution replaces expressions like \sqrt{a^2-x^2} with trig functions, making integrals with radicals easier to solve.

Frequently asked questions

When should I use Integration by Parts versus u-substitution?: Use u-substitution when the integrand is a composite function and the derivative of the inner function appears in the integrand. Use Integration by Parts when you have a product of two functions where u-substitution does not directly apply, such as $x e^x$ , $x \sin(x)$ , or $x \ln(x)$ .
How do I remember the Integration by Parts formula?: A common mnemonic is "U-dV equals U-V minus V-dU," which rhymes with the formula $\int u\,dv = uv - \int v\,du$ . Another memory aid: think of it as reversing the product rule for derivatives, which says $(uv)' = u'v + uv'$ .
What do I do if I choose $u$ and $dv$ incorrectly?: If your resulting integral $\int v\,du$ is harder than the original, you made a poor choice. Try switching your assignments or select a different $u$ entirely using the LIATE rule to guide you toward a better pairing.
Can I use Integration by Parts on $\int e^{x^2}\,dx$ ?: No. There is no elementary antiderivative for $e^{x^2}$ , so Integration by Parts cannot help you evaluate it in closed form. This integral requires numerical methods or special functions like the error function.

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Reviewed 2026-09-18